Difference between revisions of "2006 AIME I Problems/Problem 3"
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== Solution == | == Solution == | ||
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+ | Note that the product of the first <math>100</math> positive odd integers can be written as <math>1\cdot 3\cdot 5\cdot 7\cdots 195\cdot 197\cdot 199=\frac{1\cdot 2\cdots200}{2\cdot4\cdots200} = \frac{200!}{2^{100}\cdot 100!}</math> | ||
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+ | Hence, we seek the number of threes in <math>200!</math> decreased by the number of threes in <math>100!.</math> | ||
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+ | There are | ||
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+ | <math>\left\lfloor \frac{200}{3}\right\rfloor+\left\lfloor\frac{200}{9}\right\rfloor+\left\lfloor \frac{200}{27}\right\rfloor+\left\lfloor\frac{200}{81}\right\rfloor =66+22+7+2=97</math> | ||
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+ | threes in <math>200!</math> and | ||
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+ | <math>\left\lfloor \frac{100}{3}\right\rfloor+\left\lfloor\frac{100}{9}\right\rfloor+\left\lfloor \frac{100}{27}\right\rfloor+\left\lfloor\frac{100}{81}\right\rfloor=33+11+3+1=48 </math> | ||
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+ | threes in <math>100!</math> | ||
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+ | Therefore, we have a total of <math>97-48=049</math> threes. | ||
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+ | For more information, see also [[Factorial#Prime factorization| prime factorizations of a factorial]]. | ||
== See also == | == See also == |
Revision as of 14:00, 25 September 2007
Problem
Let be the product of the first
positive odd integers. Find the largest integer
such that
is divisible by
Solution
Note that the product of the first positive odd integers can be written as
Hence, we seek the number of threes in decreased by the number of threes in
There are
threes in and
threes in
Therefore, we have a total of threes.
For more information, see also prime factorizations of a factorial.
See also
2006 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |